Problem: $-10pq - 8pr + 3p + 7 = q - 1$ Solve for $p$.
Answer: Combine constant terms on the right. $-10pq - 8pr + 3p + {7} = q - {1}$ $-10pq - 8pr + 3p = q - {8}$ Notice that all the terms on the left-hand side of the equation have $p$ in them. $-10{p}q - 8{p}r + 3{p} = q - 8$ Factor out the $p$ ${p} \cdot \left( -10q - 8r + 3 \right) = q - 8$ Isolate the $p$ $p \cdot \left( -{10q - 8r + 3} \right) = q - 8$ $p = \dfrac{ q - 8 }{ -{10q - 8r + 3} }$ We can simplify this by multiplying the top and bottom by $-1$. $p= \dfrac{-q + 8}{10q + 8r - 3}$